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受欢迎的微积分 >

integral from 0 to 2 of sqrt(1+(2x)^2)

解答

\int_{0}^{2} 1 + (2 x)^{2} d x

解答

17 + \frac{1}{4} ln (4 + 17)

+1

十进制

4.64678 \dots

求解步骤

\int_{0}^{2} 1 + (2 x)^{2} d x

使用三角换元法

= \int_{0}^{a r c t a n (4)} \frac{1}{2} sec^{3} (u) d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{a r c t a n (4)} sec^{3} (u) d u

使用降次积分法

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (4)} + \frac{1}{2} \cdot \int_{0}^{a r c t a n (4)} sec (u) d u)

\int_{0}^{a r c t a n (4)} sec (u) d u = ln (4 + 17)

= \frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (4)} + \frac{1}{2} ln (4 + 17))

化简

\frac{1}{2} ([\frac{sec ^{2} ( u ) sin ( u )}{2}]_{0}^{a r c t a n (4)} + \frac{1}{2} ln (4 + 17)) : \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (4)} + \frac{1}{2} ln (4 + 17))

= \frac{1}{2} ([\frac{1}{2} sec (u) tan (u)]_{0}^{a r c t a n (4)} + \frac{1}{2} ln (4 + 17))

带入上下限计算后得:

217

= \frac{1}{2} (217 + \frac{1}{2} ln (4 + 17))

化简

= 17 + \frac{1}{4} ln (4 + 17)

作图

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